Residue fields of schemes of finite type (over $\mathbb{Z}$)

You are quite close to solving this it seems. Let me help you out with the last steps.

For 1), we may and do assume that $X$ is affine, say $X=$ Spec $A$. Let $x$ be a closed point of $X$, and let $m_x$ be the corresponding maximal ideal in $A$. Let $k(x) = A_{m_x}/m_x$ be its residue field.

Note that $A$ is a finitely generated $\mathbb Z$-algebra, and that the intersection of $m_x$ and $\mathbb Z$ is a non-zero ideal. (Indeed, suppose that $m_x \cap \mathbb Z = 0$. In this case, the point $x$ of Spec $A$ lies over the generic point of Spec $\mathbb Z$. Therefore, by the definition of a morphism of schemes, there is an induced morphism of residue fields $\mathcal O_{Spec \mathbb Z, (0)} \to A/m_x$. This corresponds to an injective morphism $\mathbb Q \to A/m_x$. (You can also see that there is such a morphism directly, of course.) But $A/m_x$ is a finitely generated $\mathbb Z$-algebra, and these can not contain $\mathbb Q$. This claim is proven in detail in the answer of Eric Wofsey) Thus, since $m_x \cap \mathbb Z$ is a non-zero prime ideal, there is a prime number $p$ (uniquely determined by $x$) such that $m_x \cap \mathbb Z = p\mathbb Z$. We sometimes say that $x$ is a point of characteristic $p$.

Since $A$ is a finitely generated $\mathbb Z$-algebra, it follows that $k(x)$ is a finitely generated $\mathbb F_p$-algebra. Since $k(x)$ is a field, it follows from Hilbert's Nullstellensatz (as you correctly say yourself) that $k(x)$ is a finite field (of characteristic $p$).

For 2), we can again assume that $X$ is affine (and of finite type over $\mathbb Z$). Let $X\subset \mathbb A^n_{\mathbb Z}$ be some closed immersion. Note that the set of (closed) points with residue field $\mathbb F_q$ is contained in the set $X(\mathbb F_q)$ of $\mathbb F_q$-rational points of $X$. It follows that $X(\mathbb F_q)$ is a subset of $\mathbb A^n(\mathbb F_q) = \mathbb F_q^n$ (as sets). The latter set is clearly finite.

As a final remark: you should think of $X$ (in the affine case) as given by finitely many polynomial equations with coefficients in $\mathbb Z$. When you look for closed points with residue field $\mathbb F_q$ on $X$ you are looking for solutions of your system of polynomial equations in $\mathbb F_q$.

In (1), the assertion that $\mathbb{Z}\cap m$ cannot be $0$ follows from the version of the Nullstellensatz which applies to arbitrary Jacobson rings (rings in which every prime is the intersection of the maximal ideals containing it), rather than just fields. This general Nullstellensatz says that if $R$ is a Jacobson ring and $A$ is a finitely-generated $R$-algebra, then $A$ is Jacobson and for any maximal ideal $m\subset A$, $m\cap R$ is maximal in $R$ and $A/m$ is a finite extension of $R/(m\cap R)$. Applying this with $R=\mathbb{Z}$ immediately gives that $\mathbb{Z}\cap m$ cannot be $0$ in your case.

Alternatively, you can deduce that $\mathbb{Z}\cap m\neq 0$ from just the Nullstellensatz for fields and the Artin-Tate lemma as follows. If $\mathbb{Z}\cap m=0$, then $A/m$ is a finitely generated $\mathbb{Q}$-algebra and hence is a finite extension of $\mathbb{Q}$ by the Nullstellensatz for $\mathbb{Q}$. By the Artin-Tate lemma applied to $\mathbb{Z}\subset\mathbb{Q}\subseteq A/m$, the fact that $A/m$ is a finitely generated $\mathbb{Z}$-algebra then implies that $\mathbb{Q}$ is also a finitely generated $\mathbb{Z}$-algebra. This is clearly false (any finitely generated subring of $\mathbb{Q}$ can have only finitely many different primes appearing in denominators).